Find $\lim_{x\to \infty}(1+5e^x)^{^{\frac{1}{x}}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $e$ (Choice C) C $e^5$ (Choice D) D The limit doesn't exist.
Taking $x$ to $\infty$ in $(1+5e^x)^{^{\frac{1}{x}}}$ results in the indeterminate form $\infty^{^{0}}$. To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=(1+5e^x)^{^{\frac{1}{x}}}$, we will find $\lim_{x\to \infty}\ln(y)$. Once we find it, we will be able to find $\lim_{x\to \infty}y$. $\ln(y) =\dfrac{\ln(1+5e^x)}{x}$ Taking $x$ to $\infty$ in $\dfrac{\ln(1+5e^x)}{x}$ results in the indeterminate form $\dfrac{\infty}{\infty}$, so now it's L'Hôpital's rule's turn to help us with our quest! $\begin{aligned} &\phantom{=}\lim_{x\to \infty}\ln(y) \\\\ &=\lim_{x\to \infty}\dfrac{\ln(1+5e^x)}{x} \\\\ &=\lim_{x\to \infty}\dfrac{\dfrac{d}{dx}[\ln(1+5e^x)]}{\dfrac{d}{dx}[x]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to \infty}\dfrac{\left(\dfrac{5e^x}{1+5e^x}\right)}{1} \\\\ &=\lim_{x\to \infty}\dfrac{5e^x}{1+5e^x} \\\\ &=1 \gray{\text{The leading term is the same}} \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to \infty}\dfrac{\dfrac{d}{dx}[\ln(1+5e^x)]}{\dfrac{d}{dx}[x]}$ actually exists. We found that $\lim_{x\to \infty}\ln(y)=1$, which means $\lim_{x\to \infty}y=e$. [Why?] In conclusion, $\lim_{x\to \infty}(1+5e^x)^{^{\frac{1}{x}}}=e$.